Introductory calculus

1. Limits

Exercises

2. Differentiation from first principles

Differentiation from first principles

Determine the derivative of

f(x)=14x−3

using first principles and find the gradient of the function at x=−5.

Answer:

f′(x)=

The gradient at x=−5 is

polynomial
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).

STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=14(x+h)−3
STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh→0f(x+h)−f(x)h=limh→01h×[f(x+h)−f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh→01h[f(x+h)−f(x)]=limh→01h[14(x+h)−3−14x−3]=limh→01h[14x+4h−3−14x−3]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.

STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (4x+4h−3) and (4x−3). We see that the two denominators do not share any factors, so the common denominator must be (4x+4h−3)×(4x−3).

f′(x)=limh→01h[4x−3−(4x+4h−3)(4x+4h−3)(4x−3)]=limh→01h[−4h(4x+4h−3)(4x−3)]
STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f′(x)=limh→0[−4(4x+4h−3)(4x−3)]=−4(4x+4(0)−3)(4x−3)=−4(4x−3)(4x−3)∴f′(x)=−4(4x−3)2
STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=−5 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f′(−5):

f′(−5)=−4(4(−5)−3)2=−4529
Submit your answer as: and

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f′(x) from first principles, if

f(x)=5x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.

  1. Which of the following is the formula used to find a derivative from first principles?

    A f′(x)=limh→0f(x)−f(x+h)h
    B f′(x)=limh→0f(x+h)−f(x)h
    C f′(x)=limh→xf(x+h)−f(x)h
    D f′(x)=limx→0f(x+h)+f(x)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f′(x)=limh→0f(x+h)−f(x)h

    So the correct formula is Option B.

    Submit your answer as:

  2. Which of the following expressions is equivalent to f(x+h)?

    A 5h2
    B 5x+5h
    C 5x2+5xh+5h2
    D 5x2+10xh+5h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=5(x+h)2.

    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=5x2

    Substituting in (x+h), we get

    f(x+h)=5(x+h)2=5(x2+2xh+h2)=5x2+10xh+5h2

    So the correct expression is Option D.

    Submit your answer as:

  3. What is the simplified expression for the numerator f(x+h)−f(x)?

    Answer: f(x+h)−f(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.

    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)−f(x)=(5x2+10xh+5h2)−(5x2)=10xh+5h2
    Submit your answer as:

  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h→0?

    The last thing to do before h→0 is to...
    A cancel h in the numerator and denominator.
    B substitute x=0.
    C simplify the numerator.
    D divide everything by 5.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f′(x)=limh→0f(x+h)−f(x)h=limh→010xh+5h2h=limh→0h(10x+5h)h=limh→0(10x+5h)

    The final thing that we need to do before evaluating the limit h→0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option A.

    Submit your answer as:

  5. What is the final answer for f′(x)?

    Answer: f′(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h→0.

    STEP: Evaluate the limit h→0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h→0

    f′(x)=limh→0(10x+5h)=10x+5(0)=10x
    Submit your answer as:

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of f′(2).

f(y)=−6y2+4y+6
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • f′(y)=
  • f′(2) =
expression
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

f′(y)=limh→0f(y+h)−f(y)h

First, determine f(y+h).

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

f′(y)=limh→0f(y+h)−f(y)h

Calculate f(y+h) for the function by substituting (y+h) for every y:

The function is: f(y)=−6y2+4y+6
∴f(y+h)=−6(y+h)2+4(y+h)+6=−6(y2+2yh+h2)+4y+4h+6=−6y2−12yh−6h2+4y+4h+6
STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

f(y+h)−f(y)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

f(y+h)−f(y)=(−6y2−12yh−6h2+4y+4h+6)−(−6y2+4y+6)=−6y2−12yh−6h2+4y+4h+6+6y2−4y−6=−6h2−12hy+4h
STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

f′(y)=limh→0f(y+h)−f(y)h=limh→0−6h2−12hy+4hh
STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

f′(y)=limh→0h(−6h−12y+4)h=limh→0(−6h−12y+4)

Now, if we apply the limit and let h=0, the expression is not undefined:

f′(y)=−6(0)−12y+4=−12y+4

Write the final answer: f′(y)=−12y+4 and f′(2)=−12(2)+4=−20

Submit your answer as: and

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Atinuke determines y′(a), the derivative of a certain function y at x=a, and arrives at the answer:

y′(a)=limh→016+h−4h

Complete the equation below for y (type in the right hand side of the equation) and determine the value of a.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • y(x)=
  • a=
expression
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

STEP: Compare y′(a) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f′(x) of a function f(x) is

f′(x)=limh→0f(x+h)−f(x)h

We can rewrite this for the function y(x), with x=a:

y′(a)=limh→0y(a+h)−y(a)h

Comparing this to the equation that we are given for y′(a), we can see that

y(a+h)=16+hy(a)=4

The first equation here shows y(x+h) evaluated at x=a. This tells us that

a+h=16+ha=16

and also that y(x)=x.

We can check this by looking at the second term, which is y(x) evaluated at x=a.

y(x)=xa=16↓y(a)=(16)=4

which is exactly the what we expect. So the correct answers are

  • y(x)=x
  • a=16
Submit your answer as: and

Exercises

Differentiation from first principles

Determine the derivative of

f(x)=1x+1

using first principles and find the gradient of the function at x=−6.

Answer:

f′(x)=

The gradient at x=−6 is

polynomial
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).

STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=1(x+h)+1
STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh→0f(x+h)−f(x)h=limh→01h×[f(x+h)−f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh→01h[f(x+h)−f(x)]=limh→01h[1(x+h)+1−1x+1]=limh→01h[1x+h+1−1x+1]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.

STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (x+h+1) and (x+1). We see that the two denominators do not share any factors, so the common denominator must be (x+h+1)×(x+1).

f′(x)=limh→01h[x+1−(x+h+1)(x+h+1)(x+1)]=limh→01h[−h(x+h+1)(x+1)]
STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f′(x)=limh→0[−1(x+h+1)(x+1)]=−1(x+(0)+1)(x+1)=−1(x+1)(x+1)∴f′(x)=−1(x+1)2
STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=−6 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f′(−6):

f′(−6)=−1((−6)+1)2=−125
Submit your answer as: and

Differentiation from first principles

Determine the derivative of

f(x)=12x−7

using first principles and find the gradient of the function at x=−5.

Answer:

f′(x)=

The gradient at x=−5 is

polynomial
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).

STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=12(x+h)−7
STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh→0f(x+h)−f(x)h=limh→01h×[f(x+h)−f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh→01h[f(x+h)−f(x)]=limh→01h[12(x+h)−7−12x−7]=limh→01h[12x+2h−7−12x−7]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.

STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (2x+2h−7) and (2x−7). We see that the two denominators do not share any factors, so the common denominator must be (2x+2h−7)×(2x−7).

f′(x)=limh→01h[2x−7−(2x+2h−7)(2x+2h−7)(2x−7)]=limh→01h[−2h(2x+2h−7)(2x−7)]
STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f′(x)=limh→0[−2(2x+2h−7)(2x−7)]=−2(2x+2(0)−7)(2x−7)=−2(2x−7)(2x−7)∴f′(x)=−2(2x−7)2
STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=−5 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f′(−5):

f′(−5)=−2(2(−5)−7)2=−2289
Submit your answer as: and

Differentiation from first principles

Determine the derivative of

f(x)=1x−9

using first principles and find the gradient of the function at x=−1.

Answer:

f′(x)=

The gradient at x=−1 is

polynomial
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).

STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=1(x+h)−9
STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh→0f(x+h)−f(x)h=limh→01h×[f(x+h)−f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh→01h[f(x+h)−f(x)]=limh→01h[1(x+h)−9−1x−9]=limh→01h[1x+h−9−1x−9]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.

STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (x+h−9) and (x−9). We see that the two denominators do not share any factors, so the common denominator must be (x+h−9)×(x−9).

f′(x)=limh→01h[x−9−(x+h−9)(x+h−9)(x−9)]=limh→01h[−h(x+h−9)(x−9)]
STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f′(x)=limh→0[−1(x+h−9)(x−9)]=−1(x+(0)−9)(x−9)=−1(x−9)(x−9)∴f′(x)=−1(x−9)2
STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=−1 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f′(−1):

f′(−1)=−1((−1)−9)2=−1100
Submit your answer as: and

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f′(x) from first principles, if

f(x)=4x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.

  1. Which of the following is the formula used to find a derivative from first principles?

    A f′(x)=limh→xf(x+h)−f(x)h
    B f′(x)=limh→0f(x)−f(x+h)h
    C f′(x)=limh→0f(x+h)−f(x)h
    D f′(x)=limx→0f(x+h)+f(x)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f′(x)=limh→0f(x+h)−f(x)h

    So the correct formula is Option C.

    Submit your answer as:

  2. Which of the following expressions is equivalent to f(x+h)?

    A 4h2
    B 4x2+8xh+4h2
    C 4x+4h
    D 4x2+16xh+4h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=4(x+h)2.

    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=4x2

    Substituting in (x+h), we get

    f(x+h)=4(x+h)2=4(x2+2xh+h2)=4x2+8xh+4h2

    So the correct expression is Option B.

    Submit your answer as:

  3. What is the simplified expression for the numerator f(x+h)−f(x)?

    Answer: f(x+h)−f(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.

    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)−f(x)=(4x2+8xh+4h2)−(4x2)=8xh+4h2
    Submit your answer as:

  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h→0?

    The last thing to do before h→0 is to...
    A cancel h in the numerator and denominator.
    B divide everything by 4.
    C substitute x=0.
    D simplify the numerator.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f′(x)=limh→0f(x+h)−f(x)h=limh→08xh+4h2h=limh→0h(8x+4h)h=limh→0(8x+4h)

    The final thing that we need to do before evaluating the limit h→0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option A.

    Submit your answer as:

  5. What is the final answer for f′(x)?

    Answer: f′(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h→0.

    STEP: Evaluate the limit h→0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h→0

    f′(x)=limh→0(8x+4h)=8x+4(0)=8x
    Submit your answer as:

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f′(x) from first principles, if

f(x)=2x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.

  1. Which of the following is the formula used to find a derivative from first principles?

    A f′(x)=limh→0f(x)−f(x+h)h
    B f′(x)=limh→xf(x+h)−f(x)h
    C f′(x)=limh→0f(x+h)−f(x)h
    D f′(x)=limx→0f(x+h)+f(x)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f′(x)=limh→0f(x+h)−f(x)h

    So the correct formula is Option C.

    Submit your answer as:

  2. Which of the following expressions is equivalent to f(x+h)?

    A 2x+2h
    B 2x2+4xh+2h2
    C 2x2+2h2
    D 2h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=2(x+h)2.

    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=2x2

    Substituting in (x+h), we get

    f(x+h)=2(x+h)2=2(x2+2xh+h2)=2x2+4xh+2h2

    So the correct expression is Option B.

    Submit your answer as:

  3. What is the simplified expression for the numerator f(x+h)−f(x)?

    Answer: f(x+h)−f(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.

    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)−f(x)=(2x2+4xh+2h2)−(2x2)=4xh+2h2
    Submit your answer as:

  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h→0?

    The last thing to do before h→0 is to...
    A divide everything by 2.
    B cancel h in the numerator and denominator.
    C substitute x=0.
    D simplify the numerator.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f′(x)=limh→0f(x+h)−f(x)h=limh→04xh+2h2h=limh→0h(4x+2h)h=limh→0(4x+2h)

    The final thing that we need to do before evaluating the limit h→0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option B.

    Submit your answer as:

  5. What is the final answer for f′(x)?

    Answer: f′(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h→0.

    STEP: Evaluate the limit h→0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h→0

    f′(x)=limh→0(4x+2h)=4x+2(0)=4x
    Submit your answer as:

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f′(x) from first principles, if

f(x)=6x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.

  1. Which of the following is the formula used to find a derivative from first principles?

    A f′(x)=limh→xf(x+h)−f(x)h
    B f′(x)=limh→0f(x+h)−f(x)h
    C f′(x)=limx→0f(x+h)+f(x)h
    D f′(x)=limh→0f(x)−f(x+h)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f′(x)=limh→0f(x+h)−f(x)h

    So the correct formula is Option B.

    Submit your answer as:

  2. Which of the following expressions is equivalent to f(x+h)?

    A 6x2+12xh+6h2
    B 6x+6h
    C 6h2
    D 6x2+6h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=6(x+h)2.

    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=6x2

    Substituting in (x+h), we get

    f(x+h)=6(x+h)2=6(x2+2xh+h2)=6x2+12xh+6h2

    So the correct expression is Option A.

    Submit your answer as:

  3. What is the simplified expression for the numerator f(x+h)−f(x)?

    Answer: f(x+h)−f(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.

    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)−f(x)=(6x2+12xh+6h2)−(6x2)=12xh+6h2
    Submit your answer as:

  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h→0?

    The last thing to do before h→0 is to...
    A cancel h in the numerator and denominator.
    B simplify the numerator.
    C substitute x=0.
    D divide everything by 6.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.

    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f′(x)=limh→0f(x+h)−f(x)h=limh→012xh+6h2h=limh→0h(12x+6h)h=limh→0(12x+6h)

    The final thing that we need to do before evaluating the limit h→0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option A.

    Submit your answer as:

  5. What is the final answer for f′(x)?

    Answer: f′(x)=
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h→0.

    STEP: Evaluate the limit h→0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h→0

    f′(x)=limh→0(12x+6h)=12x+6(0)=12x
    Submit your answer as:

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of g′(−3).

g(x)=−9x2−5x+3
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • dg(x)dx=
  • g′(−3) =
expression
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

dg(x)dx=limh→0g(x+h)−g(x)h

First, determine g(x+h).

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

dg(x)dx=limh→0g(x+h)−g(x)h

Calculate g(x+h) for the function by substituting (x+h) for every x:

The function is: g(x)=−9x2−5x+3
∴g(x+h)=−9(x+h)2−5(x+h)+3=−9(x2+2xh+h2)−5x−5h+3=−9x2−18xh−9h2−5x−5h+3
STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

g(x+h)−g(x)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

g(x+h)−g(x)=(−9x2−18xh−9h2−5x−5h+3)−(−9x2−5x+3)=−9x2−18xh−9h2−5x−5h+3+9x2+5x−3=−9h2−18hx−5h
STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

dg(x)dx=limh→0g(x+h)−g(x)h=limh→0−9h2−18hx−5hh
STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

dg(x)dx=limh→0h(−9h−18x−5)h=limh→0(−9h−18x−5)

Now, if we apply the limit and let h=0, the expression is not undefined:

dg(x)dx=−9(0)−18x−5=−18x−5

Write the final answer: dg(x)dx=−18x−5 and g′(−3)=−18(−3)+−5=49

Submit your answer as: and

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of f′(5).

f(z)=−2z2−3z−8
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • ddzf(z)=
  • f′(5) =
expression
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

ddzf(z)=limh→0f(z+h)−f(z)h

First, determine f(z+h).

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

ddzf(z)=limh→0f(z+h)−f(z)h

Calculate f(z+h) for the function by substituting (z+h) for every z:

The function is: f(z)=−2z2−3z−8
∴f(z+h)=−2(z+h)2−3(z+h)−8=−2(z2+2zh+h2)−3z−3h−8=−2z2−4zh−2h2−3z−3h−8
STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

f(z+h)−f(z)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

f(z+h)−f(z)=(−2z2−4zh−2h2−3z−3h−8)−(−2z2−3z−8)=−2z2−4zh−2h2−3z−3h−8+2z2+3z+8=−2h2−4hz−3h
STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

ddzf(z)=limh→0f(z+h)−f(z)h=limh→0−2h2−4hz−3hh
STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

ddzf(z)=limh→0h(−2h−4z−3)h=limh→0(−2h−4z−3)

Now, if we apply the limit and let h=0, the expression is not undefined:

ddzf(z)=−2(0)−4z−3=−4z−3

Write the final answer: ddzf(z)=−4z−3 and f′(5)=−4(5)+−3=−23

Submit your answer as: and

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of f′(0).

f(y)=−7y2+8y+2
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • f′(y)=
  • f′(0) =
expression
numeric
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

f′(y)=limh→0f(y+h)−f(y)h

First, determine f(y+h).

STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

f′(y)=limh→0f(y+h)−f(y)h

Calculate f(y+h) for the function by substituting (y+h) for every y:

The function is: f(y)=−7y2+8y+2
∴f(y+h)=−7(y+h)2+8(y+h)+2=−7(y2+2yh+h2)+8y+8h+2=−7y2−14yh−7h2+8y+8h+2
STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

f(y+h)−f(y)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

f(y+h)−f(y)=(−7y2−14yh−7h2+8y+8h+2)−(−7y2+8y+2)=−7y2−14yh−7h2+8y+8h+2+7y2−8y−2=−7h2−14hy+8h
STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

f′(y)=limh→0f(y+h)−f(y)h=limh→0−7h2−14hy+8hh
STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

f′(y)=limh→0h(−7h−14y+8)h=limh→0(−7h−14y+8)

Now, if we apply the limit and let h=0, the expression is not undefined:

f′(y)=−7(0)−14y+8=−14y+8

Write the final answer: f′(y)=−14y+8 and f′(0)=−14(0)+8=8

Submit your answer as: and

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Hakeema determines y′(b), the derivative of a certain function y at x=b, and arrives at the answer:

y′(b)=limh→025+h−5h

Complete the equation below for y (type in the right hand side of the equation) and determine the value of b.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • y(x)=
  • b=
expression
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

STEP: Compare y′(b) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f′(x) of a function f(x) is

f′(x)=limh→0f(x+h)−f(x)h

We can rewrite this for the function y(x), with x=b:

y′(b)=limh→0y(b+h)−y(b)h

Comparing this to the equation that we are given for y′(b), we can see that

y(b+h)=25+hy(b)=5

The first equation here shows y(x+h) evaluated at x=b. This tells us that

b+h=25+hb=25

and also that y(x)=x.

We can check this by looking at the second term, which is y(x) evaluated at x=b.

y(x)=xb=25↓y(b)=(25)=5

which is exactly the what we expect. So the correct answers are

  • y(x)=x
  • b=25
Submit your answer as: and

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Ndidi determines m′(a), the derivative of a certain function m at x=a, and arrives at the answer:

m′(a)=limh→09+h−3h

Complete the equation below for m (type in the right hand side of the equation) and determine the value of a.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • m(x)=
  • a=
expression
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

STEP: Compare m′(a) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f′(x) of a function f(x) is

f′(x)=limh→0f(x+h)−f(x)h

We can rewrite this for the function m(x), with x=a:

m′(a)=limh→0m(a+h)−m(a)h

Comparing this to the equation that we are given for m′(a), we can see that

m(a+h)=9+hm(a)=3

The first equation here shows m(x+h) evaluated at x=a. This tells us that

a+h=9+ha=9

and also that m(x)=x.

We can check this by looking at the second term, which is m(x) evaluated at x=a.

m(x)=xa=9↓m(a)=(9)=3

which is exactly the what we expect. So the correct answers are

  • m(x)=x
  • a=9
Submit your answer as: and

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Ntuthuko determines y′(a), the derivative of a certain function y at x=a, and arrives at the answer:

y′(a)=limh→09+h−3h

Complete the equation below for y (type in the right hand side of the equation) and determine the value of a.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • y(x)=
  • a=
expression
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.

STEP: Compare y′(a) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f′(x) of a function f(x) is

f′(x)=limh→0f(x+h)−f(x)h

We can rewrite this for the function y(x), with x=a:

y′(a)=limh→0y(a+h)−y(a)h

Comparing this to the equation that we are given for y′(a), we can see that

y(a+h)=9+hy(a)=3

The first equation here shows y(x+h) evaluated at x=a. This tells us that

a+h=9+ha=9

and also that y(x)=x.

We can check this by looking at the second term, which is y(x) evaluated at x=a.

y(x)=xa=9↓y(a)=(9)=3

which is exactly the what we expect. So the correct answers are

  • y(x)=x
  • a=9
Submit your answer as: and

3. Standard derivations

Finding the derivative using the rule for differentiation

Given:

g(x)=23x3−14x2+6x

Determine Dxg(x) using the rules for differentiation.

Answer: Dxg(x)=
polynomial
STEP: Recall the rule for differentiation
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn−1
STEP: Differentiate the first term
[−1 point ⇒ 2 / 3 points left]

Let's consider the first term of the function:

ddx(23x3)=3(23)x(3−1)=2x2

We follow the same rule for all the remaining terms.

STEP: Differentiate the rest of the function
[−2 points ⇒ 0 / 3 points left]
ddx(23x3−14x2+6x)=3(23)x(3−1)+2(−14)x(2−1)+1(6)x(1−1)=2x2−x2+6

Therefore, we can write the final answer:

Dxg(x)=2x2−x2+6
Submit your answer as:

Finding derivatives using the rules

Evaluate Dzg(z) using the rules for differentiation:

g(z)=−12z6+2z3+74z43
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: Dzg(z)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...−43x23=−43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in g(z)?

STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

−12z6+2z3+74z43=−12z6+2z3+74z43
STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddz(−12z6+2z3+74z43)=−(6)(12)z(6−1)+(3)(2)z(3−1)+(43)(74)z(43−1)

We write the final answer with positive exponents:

Dzg(z)=−3z5+6z2+73z13
Submit your answer as:

Finding the derivative

Calculate df(x)dx for f(x)=8x4−x2−2x0,5+4x0,25.

Answer:df(x)dx=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation is the same for decimals as it is for integers.

STEP: Focus on the rule for differentiation
[−1 point ⇒ 2 / 3 points left]

This function contains decimal values in the exponents, but the rule for differention is always the same:

f(x)=ddx(axn)=(an)xn−1For example: ddx(−3)x0,8=(0,8)(−3)x0,8−1=−2,4x−0,2
STEP: Apply the rule to the function
[−1 point ⇒ 1 / 3 points left]

For each term, multiplying the coefficient by the exponent and subtracting one from the exponent:

f′(x)=(4)(8)x4−1+(2)(−1)x2−1+(0,5)(−2)x0,5−1+(0,25)(4)x0,25−1
STEP: Simplify the answer
[−1 point ⇒ 0 / 3 points left]

Tidy up all of the calculations:

f′(x)=32x3−2x−x−0,5+x−0,75

Therefore, the final answer is:

ddx[8x4−x2−2x0,5+4x0,25]=32x3−2x−1x0,5+1x0,75
Submit your answer as:

Applying the rules for differentiation

Calculate df(y)dy for the function:

f(y)=3y4+9y+4−9y−2

Write your answer with positive exponents.

Answer: df(y)dy=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x−2 ... and ... x=x12
STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn−1
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

3y4+9y+4−9y−2=3y4+9y12+4−9y−2
STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddy(3y4+9y12+4−9y−2)=4(3)y(4−1)+12(9)y(12−1)−(−2)(9)y(−2−1)=(12)y(3)+(92)y(−12)−(−18)y(−3)=12y3+92y(−12)+18y−3

For the final answer, rewrite the derivative with positive exponents:

ddy(3y4+9y+4−9y−2)=12y3+92y12+18y3
Submit your answer as:

Using the rules to determine the derivative

Find dh(x)dx using the rules for differentiation:

h(x)=−3x2−2x
Answer: dh(x)dx=
polynomial
STEP: <no title>
[−0 points ⇒ 2 / 2 points left]

The rules for differentiation:

ddxaxn=(an)xn−1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

Let's consider the first term of the function and apply the rule:

ddx(−3x2)=2(−3)x(2−1)=−6x1

We repeat this process for each of the remaining terms that make up the given function.

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Therefore, the final answer is:

dh(x)dx=−6x−2

Important: the derivative of a constant term, such as −4, will always be zero.

We can apply the rule to see why this is true:

ddx[−4]=ddx[−4.x0]=0.[−4.x0−1]⟵ exponent (zero)multiply by the=0.[−4x]=0
Submit your answer as:

Using the rules for differentiation

Given k(z)=1z5+1z−3z−3, determine ddzk(z).

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: ddzk(z)=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.

STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation:

ddx(axn)=(an)xn−1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.

STEP: Move the variable from the denominator
[−1 point ⇒ 3 / 4 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

1z5+1z−3z−3=1z5+1z1−3z−3
STEP: Differentiate the function
[−2 points ⇒ 1 / 4 points left]

Apply the rule for differentiation:

ddzk(z)=ddz(1z5+1z1−3z−3)=(5)(1)z(5−1)+(1)(1)z(1−1)−(−3)(3)z(−3−1)=(5)z4+(1)z0+(9)z−4
STEP: Rearrange the terms
[−1 point ⇒ 0 / 4 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

ddzk(z)=5z4+1+9z4
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!
Submit your answer as:

Exercises

Finding the derivative using the rule for differentiation

Given:

p(z)=−4z3−3z2+34z−2

Determine ddzp(z) using the rules for differentiation.

Answer: ddzp(z)=
polynomial
STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn−1
STEP: Differentiate the first term
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function:

ddz(−4z3)=3(−4)z(3−1)=−12z2

We follow the same rule for all the remaining terms.

STEP: Differentiate the rest of the function
[−3 points ⇒ 0 / 4 points left]
ddz(−4z3−3z2+34z−2)=3(−4)z(3−1)+2(−3)z(2−1)+1(34)z(1−1)+0(−2)z(0−1)=−12z2−6z+34

Therefore, we can write the final answer:

ddzp(z)=−12z2−6z+34
Submit your answer as:

Finding the derivative using the rule for differentiation

Given:

p=−3x4+34x2−14x−2

Determine Dxp using the rules for differentiation.

Answer: Dxp=
polynomial
STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn−1
STEP: Differentiate the first term
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function:

ddx(−3x4)=4(−3)x(4−1)=−12x3

We follow the same rule for all the remaining terms.

STEP: Differentiate the rest of the function
[−3 points ⇒ 0 / 4 points left]
ddx(−3x4+34x2−14x−2)=4(−3)x(4−1)+2(34)x(2−1)+1(−14)x(1−1)+0(−2)x(0−1)=−12x3+3x2−14

Therefore, we can write the final answer:

Dxp=−12x3+3x2−14
Submit your answer as:

Finding the derivative using the rule for differentiation

Given:

f=−y2+3

Determine ddyf using the rules for differentiation.

Answer: ddyf=
polynomial
STEP: Recall the rule for differentiation
[−0 points ⇒ 2 / 2 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn−1
STEP: Differentiate the first term
[−1 point ⇒ 1 / 2 points left]

Let's consider the first term of the function:

ddy(−1y2)=2(−1)y(2−1)=−2y1

We follow the same rule for all the remaining terms.

STEP: Differentiate the rest of the function
[−1 point ⇒ 0 / 2 points left]
ddy(−y2+3)=2(−1)y(2−1)+0(3)y(0−1)=−2y

Therefore, we can write the final answer:

ddyf=−2y
Submit your answer as:

Finding derivatives using the rules

Evaluate m′(y) using the rules for differentiation:

m(y)=14y5+y3−2y
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: m′(y)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...−43x23=−43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in m(y)?

STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

14y5+y3−2y=14y5+y32−2y12
STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddy(14y5+y32−2y12)=(5)(14)y(5−1)+(32)y(32−1)−(12)(2)y(12−1)

We write the final answer with positive exponents:

m′(y)=54y4+32y12−1y12
Submit your answer as:

Finding derivatives using the rules

Evaluate dn(x)dx using the rules for differentiation:

n(x)=x72+3x3+4x54
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: dn(x)dx=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...−43x23=−43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in n(x)?

STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

x72+3x3+4x54=x72+3x3+4x54
STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddx(x72+3x3+4x54)=(72)x(72−1)+(3)(3)x(3−1)+(54)(4)x(54−1)

We write the final answer with positive exponents:

dn(x)dx=72x52+9x2+5x14
Submit your answer as:

Finding derivatives using the rules

Evaluate ddxf(x) using the rules for differentiation:

f(x)=x7−3x2−32x3
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: ddxf(x)=
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...−43x23=−43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in f(x)?

STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

x7−3x2−32x3=x7−3x2−32x32
STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddx(x7−3x2−32x32)=(7)x(7−1)−(2)(3)x(2−1)−(32)(32)x(32−1)

We write the final answer with positive exponents:

ddxf(x)=7x6−6x−94x12
Submit your answer as:

Finding the derivative

Calculate ddzf(z) for f(z)=3z3−7z0,8−4z0,4.

Answer:ddzf(z)=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation is the same for decimals as it is for integers.

STEP: Focus on the rule for differentiation
[−1 point ⇒ 2 / 3 points left]

This function contains decimal values in the exponents, but the rule for differention is always the same:

f(z)=ddz(azn)=(an)zn−1For example: ddz(−3)z0,8=(0,8)(−3)z0,8−1=−2,4z−0,2
STEP: Apply the rule to the function
[−1 point ⇒ 1 / 3 points left]

For each term, multiplying the coefficient by the exponent and subtracting one from the exponent:

f′(z)=(3)(3)z3−1+(0,8)(−7)z0,8−1+(0,4)(−4)z0,4−1
STEP: Simplify the answer
[−1 point ⇒ 0 / 3 points left]

Tidy up all of the calculations:

f′(z)=9z2−5,6z−0,2−1,6z−0,6

Therefore, the final answer is:

ddz[3z3−7z0,8−4z0,4]=9z2−5,6z0,2−1,6z0,6
Submit your answer as:

Finding the derivative

Calculate dh(x)dx for h(x)=(−3x2+1)(−3x2+2).

Answer:dh(x)dx=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Simplify the expression by expanding the brackets and gathering up like terms.

STEP: Expand the function into a polynomial
[−1 point ⇒ 2 / 3 points left]

First expand the brackets:

h(x)=(−3x2)(−3x2)+(−3x2)(1)+(2)(−3x2)+(2)(1)
STEP: Simplify the expression
[−1 point ⇒ 1 / 3 points left]

Now simplify as much as possible:

h(x)=9x4−9x2+2
STEP: Evaluate the derivative using the rule for differentiation
[−1 point ⇒ 0 / 3 points left]

Use the rule to find the derivative:

h′(x)=(4)(9)x(4−1)−(2)(9)x(2−1)+0

Therefore, the final answer is:

ddx[(−3x2+1)(−3x2+2)]=36x3−18x
Submit your answer as:

Finding the derivative

Calculate g′(z) for g(z)=12z2(3z5−6z4+3z2−2).

Answer:g′(z)=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Begin by breaking the large fraction into smaller fractions.

STEP: Break the fraction down into separate fractions
[−1 point ⇒ 2 / 3 points left]

Begin by writing the function as a group of fractions, each with the same denominator:

g(z)=3z52z2−6z42z2+3z22z2−22z2
STEP: Simplify each term
[−1 point ⇒ 1 / 3 points left]

Simplify each of the terms as much as possible:

g(z)=3z32−3z2+32−1z2
Use negative exponents to rewrite any fractions with variables in the denominator:
g(z)=3z32−3z2+32−1z−2
STEP: Evaluate the derivative using the rule for differentiation
[−1 point ⇒ 0 / 3 points left]

Now evaluate the derivative with the rule.

g(z)=(3)(32)z(3−1)−(2)(3)z(2−1)+0−(−2)(1)z(−2−1)=92z2−6z1+2z−3

Therefore, the final answer is:

ddz[12z2(3z5−6z4+3z2−2)]=9z22−6z+2z3
Submit your answer as:

Applying the rules for differentiation

Calculate j′(x) for the function:

j(x)=−8x2−3x3−5x−3

Write your answer with positive exponents.

Answer: j′(x)=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x−2 ... and ... x=x12
STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn−1
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

−8x2−3x3−5x−3=−8x2−3x13−5x−3
STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddx(−8x2−3x13−5x−3)=2(−8)x(2−1)−13(3)x(13−1)−(−3)(5)x(−3−1)=(−16)x(1)−(1)x(−23)−(−15)x(−4)=−16x1−1x(−23)+15x−4

For the final answer, rewrite the derivative with positive exponents:

ddx(−8x2−3x3−5x−3)=−16x−1x23+15x4
Submit your answer as:

Applying the rules for differentiation

Calculate Dym(y) for the function:

m(y)=−4y4+3y12+1+9y

Write your answer with positive exponents.

Answer: Dym(y)=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x−2 ... and ... x=x12
STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn−1
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

−4y4+3y12+1+9y=−4y4+3y12+1+9y(−1)
STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddy(−4y4+3y12+1+9y(−1))=4(−4)y(4−1)+12(3)y(12−1)+(−1)(9)y(−1−1)=(−16)y(3)+(32)y(−12)+(−9)y(−2)=−16y3+32y(−12)−9y−2

For the final answer, rewrite the derivative with positive exponents:

ddy(−4y4+3y12+1+9y)=−16y3+32y12−9y2
Submit your answer as:

Applying the rules for differentiation

Calculate ddzj(z) for the function:

j(z)=−9z2−8z+9+7z−3

Write your answer with positive exponents.

Answer: ddzj(z)=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x−2 ... and ... x=x12
STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn−1
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

−9z2−8z+9+7z−3=−9z2−8z12+9+7z−3
STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddz(−9z2−8z12+9+7z−3)=2(−9)z(2−1)−12(8)z(12−1)+(−3)(7)z(−3−1)=(−18)z(1)−(4)z(−12)+(−21)z(−4)=−18z1−4z(−12)−21z−4

For the final answer, rewrite the derivative with positive exponents:

ddz(−9z2−8z+9+7z−3)=−18z−4z12−21z4
Submit your answer as:

Using the rules to determine the derivative

Find ddxf(x) using the rules for differentiation:

f(x)=2x3+x2+5x−5
Answer: ddxf(x)=
polynomial
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The rules for differentiation:

ddxaxn=(an)xn−1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function and apply the rule:

ddx(2x3)=3(2)x(3−1)=6x2

We repeat this process for each of the remaining terms that make up the given function.

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Therefore, the final answer is:

ddxf(x)=6x2+2x+5

Important: the derivative of a constant term, such as 2, will always be zero.

We can apply the rule to see why this is true:

ddx[2]=ddx[2.x0]=0.[2.x0−1]⟵ exponent (zero)multiply by the=0.[2x]=0
Submit your answer as:

Using the rules to determine the derivative

Find dh(x)dx using the rules for differentiation:

h(x)=6x3+3x2+5x+4
Answer: dh(x)dx=
polynomial
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The rules for differentiation:

ddxaxn=(an)xn−1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function and apply the rule:

ddx(6x3)=3(6)x(3−1)=18x2

We repeat this process for each of the remaining terms that make up the given function.

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Therefore, the final answer is:

dh(x)dx=18x2+6x+5

Important: the derivative of a constant term, such as −3, will always be zero.

We can apply the rule to see why this is true:

ddx[−3]=ddx[−3.x0]=0.[−3.x0−1]⟵ exponent (zero)multiply by the=0.[−3x]=0
Submit your answer as:

Using the rules to determine the derivative

Find df(x)dx using the rules for differentiation:

f(x)=−5x3+2x2+4x−6
Answer: df(x)dx=
polynomial
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The rules for differentiation:

ddxaxn=(an)xn−1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.

STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function and apply the rule:

ddx(−5x3)=3(−5)x(3−1)=−15x2

We repeat this process for each of the remaining terms that make up the given function.

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Therefore, the final answer is:

df(x)dx=−15x2+4x+4

Important: the derivative of a constant term, such as 3, will always be zero.

We can apply the rule to see why this is true:

ddx[3]=ddx[3.x0]=0.[3.x0−1]⟵ exponent (zero)multiply by the=0.[3x]=0
Submit your answer as:

Using the rules for differentiation

Given j(x)=4x4+53x3−3x5+6x−2, determine ddxj(x).

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: ddxj(x)=
polynomial
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.

STEP: Recall the rule for differentiation
[−0 points ⇒ 5 / 5 points left]

The rule for differentiation:

ddx(axn)=(an)xn−1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.

STEP: Move the variable from the denominator
[−1 point ⇒ 4 / 5 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

4x4+53x3−3x5+6x−2=4x−4+53x−3−3x−5+6x−2
STEP: Differentiate the function
[−3 points ⇒ 1 / 5 points left]

Apply the rule for differentiation:

ddxj(x)=ddx(4x−4+53x−3−3x−5+6x−2)=(−4)(4)x(−4−1)+(−3)(53)x(−3−1)−(−5)(3)x(−5−1)+(−2)(6)x(−2−1)=−(16)x−5−(5)x−4+(15)x−6−(12)x−3
STEP: Rearrange the terms
[−1 point ⇒ 0 / 5 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

ddxj(x)=−12x3−5x4−16x5+15x6
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!
Submit your answer as:

Using the rules for differentiation

Given h(z)=1z4+1z3+12z−1, determine ddzh(z).

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: ddzh(z)=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.

STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation:

ddx(axn)=(an)xn−1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.

STEP: Move the variable from the denominator
[−1 point ⇒ 3 / 4 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

1z4+1z3+12z−1=1z−4+1z3+12z−1
STEP: Differentiate the function
[−2 points ⇒ 1 / 4 points left]

Apply the rule for differentiation:

ddzh(z)=ddz(1z−4+1z3+12z−1)=(−4)(1)z(−4−1)+(3)(1)z(3−1)+(−1)(12)z(−1−1)=−(4)z−5+(3)z2−(12)z−2
STEP: Rearrange the terms
[−1 point ⇒ 0 / 4 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

ddzh(z)=3z2−12z2−4z5
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!
Submit your answer as:

Using the rules for differentiation

Given k(z)=1z3+2z−z−2, determine dk(z)dz.

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: dk(z)dz=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.

STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation:

ddx(axn)=(an)xn−1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.

STEP: Move the variable from the denominator
[−1 point ⇒ 3 / 4 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

1z3+2z−z−2=1z−3+2z−1−z−2
STEP: Differentiate the function
[−2 points ⇒ 1 / 4 points left]

Apply the rule for differentiation:

ddzk(z)=ddz(1z−3+2z−1−z−2)=(−3)(1)z(−3−1)+(−1)(2)z(−1−1)−(−2)z(−2−1)=−(3)z−4−(2)z−2+2z−3
STEP: Rearrange the terms
[−1 point ⇒ 0 / 4 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

dk(z)dz=−2z2+2z3−3z4
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!
Submit your answer as:

4. Rules of differentiation

Exercises

5. Practical applications

Rate of change: water in a lake

A farmer in the Eastern Cape has a lake on his property. The amount of water in the lake is controlled by a powerful pump. The farmer's daughter calculates that the volume of water in the lake is summarised by this formula:

V(d)=−3d2+47d+16

where V is the volume of water in kilolitres and d represents the number of days.

  1. What is the rate of change of the volume in the lake after 11 days?

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.

    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a lake and the amount of water is changing. There is a pump pushing water into or out of the lake as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the lake.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=−3d2+47d+16∴V′(d)=−6d+47
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=11 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=11.

    V′(d)=−6d+47V′(11)=−6(11)+47=−66+47=−19

    After 11 days, the rate of change of volume of the water in the lake is -19 kilolitres per day.

    Submit your answer as:

  2. At the end of 11 days, is the volume of water in the lake increasing or decreasing? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.

    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the lake (the volume) is increasing or decreasing when d=11 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: decreasing because the rate of change is negative.

    Submit your answer as:

  3. The volume is at a maximum after 476 days. Determine the maximum volume.

    Answer: Vmax= cm3.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Do we use the volume function or the derivative to find the answer to this question?

    STEP: Substitute the given value of d into the volume function
    [−2 points ⇒ 0 / 2 points left]

    We need to calculate the maximum volume according to the function V(d). The question tells us that the volume is at its maximum value when d=476 days; and the volume formula tells us the volume of water in the lake at time d. To find the maximum volume, substitute d=476 into the function and evaluate.

    V(d)=−3d2+47d+16V(476)=−3(476)2+47(476)+16Vmax=−220912+22096+16=240112

    The calculation shows that after 476 days, the lake has a maximum volume of 240112 kilolitres of water.

    Therefore, the lake has a maximum volume of 240112 (≈200,08) kilolitres of water.

    Submit your answer as:

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 30°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3

  1. Determine r in terms of h.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: r= cm
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

    STEP: Find r in terms of h
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find r in terms of h:

    hr=tan⁡(30°)=13∴r=3hcm
    Submit your answer as:

  2. Determine the rate at which volume changes with h when h is equal to 5 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The rate of change is cm3/cm
    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of h.

    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h
    STEP: Substitute r in terms of h into the formula
    [−2 points ⇒ 2 / 5 points left]

    We can calculate the rate of change by finding the derivative. Since we want to find the derivative of V with respect to h, we need to have the entire formula in terms of h.

    V=13πr2h=13π(3h)2h=13π(3h2)h=πh3
    STEP: Determine the derivative of V with respect to h
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to h:

    dVdh=3πh2

    We can now evaluate this at h=5 cm.

    dVdh|h=5=3π(5)2=235,61944...≈235,62 cm3/cm
    Submit your answer as:

Calculus applications: optimisation

Mphikeleli teaches Business Studies in Gauteng. He wants to make fake money so that his students can use it to learn about business transactions. Mphikeleli decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of x, as labelled. The height of the rectangle will be 3x and the width is 9 cm.

  1. Find an expression for the shaded area of the fake money in terms of x.

    Answer: Shaded area A=
    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of x ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(3x)=3x2.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(9)×(3x)=(x)2=π(3x2)2=27x=x2=9π4x2
    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable x. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(x)shaded=Arectangle−2×Asquare−Acircle=27x−2(x2)−9π4x2=27x−2x2−9π4x2

    It is common to move the quadratic terms of x to the beginning of the function; so the shaded area is: A(x)=−9π4x2−2x2+27x.

    Submit your answer as:

  2. Determine the value of x for which the shaded area of the fake money will be maximised. Round your answer to 3 decimal places.

    Answer: x≈ cm
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(x)=−9π4x2−2x2+27x=−(2+9π4)x2+27x is a quadratic function. Notice that the quadratic coefficient, −(2+9π4), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which x value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(x)=−9π4x2−2x2+27xA′(x)=−9π2x−4x+27
    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=−9π2x−4x+270=−(4+9π2)x+27⟵take out the x0=−(18,137...)x+27(18,137...)x=27x=27(18,137...)x≈1,489

    The value of x which maximizes the shaded area (rounded to 3 decimal places) is 1,489 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

    Submit your answer as:

  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 3 decimal places.

    Answer: height ≈ cm
    one-of
    type(numeric.noerror)
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 3x. We want the height which corresponds with the largest shaded area, so we need to use the value of x which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 3x. We also know from question (2) that the shaded area will be maximised when x=1,489. Therefore, we need to substitute this value into the expression for the height:

    heightmax=3x≈3(1,489)≈4,467

    When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be 4,467 cm.

    Submit your answer as:

Maximising the product of two unknown numbers

Two positive numbers have a sum of 20. One of the numbers is multiplied by the cube of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.

STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 20, and we are multiplying one of the numbers by the cube of the other number.

Start by picking symbols for the two numbers. We will use x and y. Now use these variables to write equations:

The sum of the numbers is 20:

x+y=20…Eqn 1

One number multiplied by the cube of the other:

P=xy3…Eqn 2

We use P to represent the product.

STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely x and y. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make x the subject and then substitute this expression into equation 2:

x+y=20∴x=20−y

Substituting:

P=xy3=(20−y)y3=20y3−y4=−y4+20y3

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.

STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=−y4+20y3∴dPdy=−4y3+60y2
STEP: Find the value of y which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for y.

dPdy=−4y3+60y2(0)=−4y3+60y2

Divide through by a factor of −4:

0=y3−15y20=y2(y−15)y=0ory=15

The solution y=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, y=15 corresponds to the maximum product.

STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of x which corresponds to this maximum, and then use both values to find the maximum product.

x=20−y=20−(15)=5

Finally:

P=xy3∴Pmax=(5)(15)3=(5)(3375)=16875

Therefore, the maximum value of the product is 16875.

NOTE:There is a slightly different way to get the answer; once we have the value y=15, we can substitute it into the equation P=−y4+20y3 to get the maximum product. If we solve the question that way, we do not need to find the value of x at the maximum like we did above.
Submit your answer as:

Calculus applications: optimisation

The Good Eats food company makes soup powder and the company wants to package the powder in small boxes, like the one shown below. The company will put 48 cubic centimetres of their food product into each box. The box has a base with a width of 5x cm, a depth of 2x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 5x cm, 2x cm, and h cm.

    Vbox=width×depth×height=(5x)(2x)(h)=10hx2

    Therefore, the volume function for the box is 10hx2.

    Submit your answer as:

  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(5x)(h)+2(2x)(h)+2(5x)(2x)=14hx+20x2

    Therefore, the surface area equation for the box is 14hx+20x2.

    Submit your answer as:

  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 48 cm3. What can you do with that information?

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=10hx2 and surface area =14hx+20x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 48 cm3. Therefore, the equation for volume can be written as: 48=10hx2.

    Solve for h:

    48=10hx24810x2=h⇒h=245x2

    We have found that h=245x2.

    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=14hx+20x2=14(245x2)x+20x2=3365x+20x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 20x2+3365x.

    Submit your answer as:

  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.

    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=20x2+3365x.

    It looks like the minimum surface area is close to 84 or 85. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=20x2+3365xddx[S.A.(x)]=40x−3365x2
    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=40x−3365x2above by x2 to get this:multiply the equation→0=40x3−33653365=40x3336540=x3∴x=1,1887...

    The value of x which gives the minimum surface area is xmin≈1,19. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=20x2+3365x=20(1,1887...)2+3365(1,1887...)=20(1,4132...)+56,5283...=84,7924...2placesround to→≈84,79

    The minimum surface area for the box is 84,79 cm2. This answer also agrees with the graph above.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=245x2 and we also know that the width is simply w=5x. Use these two formulae to calculate the minimum values for each quantity:

    h=245x2w=5xhmin=245(1,1887...)2wmin=5(1,1887...)=3,3965...=5,9439...≈3,4≈5,94

    The height and width corresponding to the minimum surface area are h= 3,4 cm and w= 5,94 cm.

    Submit your answer as: andand

  5. If the cardboard used to make the boxes costs R0,005 per square centimetre, how much will it cost the company to produce 2 400 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 2 400 boxes is R .
    numeric
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 84,79 cm2 and the cardboard used to make the box costs R0,005 per cm2. How can you use these numbers to find the cost of one box?

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 2 400 boxes if each box requires 84,79 cm2 of cardboard. Given that that material for the boxes costs R0,005 per cm2, the total cost will be:

    Cost=(2 400 boxes)(R0,005cm2)(84,79 cm2box)=R1 017,48

    Therefore, the total cost for 2 400 boxes is R1 017,48.

    Submit your answer as:

Rate of change: drug in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain drug in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=−t3+9t2+64t

where 0≤t≤13.

  1. Determine the number of molecules of the drug in the bloodstream 4 minutes after the drug was taken.

    Answer:

    The number of molecules of the drug in the blood stream 4 minutes after the drug was taken is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 4, and evaluate.

    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 4, and evaluate.

    M(t)=−t3+9t2+64tM(4)=−(4)3+9(4)2+64(4)=336

    The number of molecules of the drug in the bloodstream 4 minutes after the drug was taken is 336.

    Submit your answer as:

  2. Determine the rate at which the number of molecules of the drug in the bloodstream is changing exactly 2 minutes after the drug was taken.

    Answer:

    The rate at which the number of molecules of the drug in the bloodstream is changing exactly 2 minutes after the drug was taken is molecules per minute.

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.

    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=−t3+9t2+64tM′(t)=−3t2+18t+64
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M′(t)=−3t2+18t+64M′(2)=−3(2)2+18(2)+64=88

    The rate at which the number of molecules of the drug in the bloodstream is changing exactly 2 minutes after the drug was taken is 88 molecules per minute.

    Submit your answer as:

  3. How many minutes after taking the drug will the rate at which the number of molecules of the drug in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum after minute(s).

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.

    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M′(t)=−3t2+18t+64M′′(t)=−6t+18
    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M′′(t)=−6t+180=−6t+183=t

    The maximum rate of change of the number of molecules of the drug in the bloodstream is after 3 minutes.

    Submit your answer as:

Exercises

Rate of change: water in a water reservoir

On a farm in the Eastern Cape there is a water reservoir. There is a large pump which controls the amount of water in the water reservoir. The farmer is a mathematician, and he calculates that the volume of water in the water reservoir is given by the formula:

V(d)=−2d2+19d+10

where V is the volume of water in kilolitres and d represents the number of days.

  1. Determine the rate of change of the volume of the water reservoir with respect to time after 4 days.

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.

    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a water reservoir and the amount of water is changing. There is a pump pushing water into or out of the water reservoir as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the water reservoir.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=−2d2+19d+10∴V′(d)=−4d+19
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=4 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=4.

    V′(d)=−4d+19V′(4)=−4(4)+19=−16+19=3

    After 4 days, the rate of change of volume of the water in the water reservoir is 3 kilolitres per day.

    Submit your answer as:

  2. At the end of 4 days, is the volume of water in the water reservoir increasing or decreasing? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.

    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the water reservoir (the volume) is increasing or decreasing when d=4 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: increasing because the rate of change is positive.

    Submit your answer as:

  3. After how many days will the water reservoir be empty?

    Answer: It will be empty after d= days.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do we use the volume function or the derivative to find the answer to this question?

    STEP: Set the volume equal to zero and solve for d
    [−3 points ⇒ 0 / 3 points left]

    We must figure out when the water reservoir will be empty. The words "the water reservoir is empty" means that the volume is zero. Therefore, we should substitute V=0 into the volume function and then solve for d. (The solution below uses factorisation, but we could also use the quadratic formula.)

    V(d)=−2d2+19d+100=−2d2+19d+100=2d2−19d−10(makes d2 coefficient positive )0=(2d+1)(d−10)∴d=−12ord=10

    We are calculating when the water reservoir is empty: the answer cannot be negative because time (the number of days) cannot be negative. Therefore we should take only the positive answer.

    Therefore, the water reservoir will be empty after 10 days.

    Submit your answer as:

Rate of change: water in a water reservoir

On a farm in the Northern Cape there is a water reservoir. There is a large pump which controls the amount of water in the water reservoir. The farmer is a mathematician, and he calculates that the volume of water in the water reservoir is given by the formula:

V(d)=−3d2+43d+30

where V is the volume of water in kilolitres and d represents the number of days.

  1. What is the rate of change of the volume in the water reservoir after 2 days?

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.

    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a water reservoir and the amount of water is changing. There is a pump pushing water into or out of the water reservoir as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the water reservoir.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=−3d2+43d+30∴V′(d)=−6d+43
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=2 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=2.

    V′(d)=−6d+43V′(2)=−6(2)+43=−12+43=31

    After 2 days, the rate of change of volume of the water in the water reservoir is 31 kilolitres per day.

    Submit your answer as:

  2. Is the volume of water increasing or decreasing at the end of 2 days? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.

    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the water reservoir (the volume) is increasing or decreasing when d=2 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: increasing because the rate of change is positive.

    Submit your answer as:

  3. The volume is at a maximum after 436 days. What is the maximum volume of water in the water reservoir?

    Answer: Vmax= cm3.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Do we use the volume function or the derivative to find the answer to this question?

    STEP: Substitute the given value of d into the volume function
    [−2 points ⇒ 0 / 2 points left]

    We need to calculate the maximum volume according to the function V(d). The question tells us that the volume is at its maximum value when d=436 days; and the volume formula tells us the volume of water in the water reservoir at time d. To find the maximum volume, substitute d=436 into the function and evaluate.

    V(d)=−3d2+43d+30V(436)=−3(436)2+43(436)+30Vmax=−184912+18496+30=220912

    The calculation shows that after 436 days, the water reservoir has a maximum volume of 220912 kilolitres of water.

    Therefore, the water reservoir has a maximum volume of 220912 (≈184,08) kilolitres of water.

    Submit your answer as:

Rate of change: water in a dam

On a farm in the Western Cape there is a dam. There is a large pump which controls the amount of water in the dam. The farmer is a mathematician, and he calculates that the volume of water in the dam is given by the formula:

V(d)=−3d2+49d+90

where V is the volume of water in kilolitres and d represents the number of days.

  1. What is the rate of change of the volume in the dam after 15 days?

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.

    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a dam and the amount of water is changing. There is a pump pushing water into or out of the dam as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the dam.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=−3d2+49d+90∴V′(d)=−6d+49
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=15 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=15.

    V′(d)=−6d+49V′(15)=−6(15)+49=−90+49=−41

    After 15 days, the rate of change of volume of the water in the dam is -41 kilolitres per day.

    Submit your answer as:

  2. At the end of 15 days, is the volume of water in the dam increasing or decreasing? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.

    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the dam (the volume) is increasing or decreasing when d=15 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: decreasing because the rate of change is negative.

    Submit your answer as:

  3. The volume is at a maximum after 496 days. Determine the maximum volume.

    Answer: Vmax= cm3.
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Do we use the volume function or the derivative to find the answer to this question?

    STEP: Substitute the given value of d into the volume function
    [−2 points ⇒ 0 / 2 points left]

    We need to calculate the maximum volume according to the function V(d). The question tells us that the volume is at its maximum value when d=496 days; and the volume formula tells us the volume of water in the dam at time d. To find the maximum volume, substitute d=496 into the function and evaluate.

    V(d)=−3d2+49d+90V(496)=−3(496)2+49(496)+90Vmax=−240112+24016+90=348112

    The calculation shows that after 496 days, the dam has a maximum volume of 348112 kilolitres of water.

    Therefore, the dam has a maximum volume of 348112 (≈290,08) kilolitres of water.

    Submit your answer as:

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 30°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3

  1. Determine r in terms of h.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: r= cm
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

    STEP: Find r in terms of h
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find r in terms of h:

    hr=tan⁡(30°)=13∴r=3hcm
    Submit your answer as:

  2. Determine the derivative of the volume of water with respect to h when h is equal to 9 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The derivative of the volume is cm3/cm
    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of h.

    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h
    STEP: Substitute r in terms of h into the formula
    [−2 points ⇒ 2 / 5 points left]

    Since we want to find the derivative of V with respect to h, we need to have the entire formula in terms of h.

    V=13πr2h=13π(3h)2h=13π(3h2)h=πh3
    STEP: Determine the derivative of V with respect to h
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to h:

    dVdh=3πh2

    We can now evaluate this at h=9 cm.

    dVdh|h=9=3π(9)2=763,40701...≈763,41 cm3/cm
    Submit your answer as:

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 30°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3

  1. Determine r in terms of h.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: r= cm
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

    STEP: Find r in terms of h
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find r in terms of h:

    hr=tan⁡(30°)=13∴r=3hcm
    Submit your answer as:

  2. Determine the derivative of the volume of water with respect to h when h is equal to 6 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The derivative of the volume is cm3/cm
    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of h.

    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h
    STEP: Substitute r in terms of h into the formula
    [−2 points ⇒ 2 / 5 points left]

    Since we want to find the derivative of V with respect to h, we need to have the entire formula in terms of h.

    V=13πr2h=13π(3h)2h=13π(3h2)h=πh3
    STEP: Determine the derivative of V with respect to h
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to h:

    dVdh=3πh2

    We can now evaluate this at h=6 cm.

    dVdh|h=6=3π(6)2=339,29200...≈339,29 cm3/cm
    Submit your answer as:

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 60°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3

  1. Determine r in terms of h.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: r= cm
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.

    STEP: Find r in terms of h
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find r in terms of h:

    hr=tan⁡(60°)=3∴r=h3cm
    Submit your answer as:

  2. Determine the rate at which volume changes with h when h is equal to 9 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The rate of change is cm3/cm
    one-of
    type(numeric.abserror(0.005))
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of h.

    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h
    STEP: Substitute r in terms of h into the formula
    [−2 points ⇒ 2 / 5 points left]

    We can calculate the rate of change by finding the derivative. Since we want to find the derivative of V with respect to h, we need to have the entire formula in terms of h.

    V=13πr2h=13π(h3)2h=13π(h23)h=19πh3
    STEP: Determine the derivative of V with respect to h
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to h:

    dVdh=319πh3=13πh2

    We can now evaluate this at h=9 cm.

    dVdh|h=9=13π(9)2=84,82300...≈84,82 cm3/cm
    Submit your answer as:

Calculus applications: optimisation

Nthofeela teaches Business Studies in KwaZulu-Natal. He wants to make fake money so that his students can use it to learn about business transactions. Nthofeela decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of y, as labelled. The height of the rectangle will be 4y and the width is 6 cm.

  1. Find an expression for the shaded area of the fake money in terms of y.

    Answer: Shaded area A=
    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of y ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(4y)=2y.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(6)×(4y)=(y)2=π(2y)2=24y=y2=4πy2
    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable y. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(y)shaded=Arectangle−2×Asquare−Acircle=24y−2(y2)−4πy2=24y−2y2−4πy2

    It is common to move the quadratic terms of y to the beginning of the function; so the shaded area is: A(y)=−4πy2−2y2+24y.

    Submit your answer as:

  2. Determine the value of y for which the shaded area of the fake money will be maximised. Round your answer to 3 decimal places.

    Answer: y≈ cm
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(y)=−4πy2−2y2+24y=−(2+4π)y2+24y is a quadratic function. Notice that the quadratic coefficient, −(2+4π), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which y value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(y)=−4πy2−2y2+24yA′(y)=−8πy−4y+24
    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=−8πy−4y+240=−(4+8π)y+24⟵take out the y0=−(29,132...)y+24(29,132...)y=24y=24(29,132...)y≈0,824

    The value of y which maximizes the shaded area (rounded to 3 decimal places) is 0,824 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

    Submit your answer as:

  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 3 decimal places.

    Answer: height ≈ cm
    one-of
    type(numeric.noerror)
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 4y. We want the height which corresponds with the largest shaded area, so we need to use the value of y which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 4y. We also know from question (2) that the shaded area will be maximised when y=0,824. Therefore, we need to substitute this value into the expression for the height:

    heightmax=4y≈4(0,824)≈3,296

    When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be 3,296 cm.

    Submit your answer as:

Calculus applications: optimisation

Chibueze teaches Business Studies in the Free State. He wants to make fake money so that his students can use it to learn about business transactions. Chibueze decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of d, as labelled. The height of the rectangle will be 6d and the width is 7 cm.

  1. Find an expression for the shaded area of the fake money in terms of d.

    Answer: Shaded area A=
    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of d ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(6d)=3d.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(7)×(6d)=(d)2=π(3d)2=42d=d2=9πd2
    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable d. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(d)shaded=Arectangle−2×Asquare−Acircle=42d−2(d2)−9πd2=42d−2d2−9πd2

    It is common to move the quadratic terms of d to the beginning of the function; so the shaded area is: A(d)=−9πd2−2d2+42d.

    Submit your answer as:

  2. Determine the value of d for which the shaded area of the fake money will be maximised. Round your answer to 2 decimal places.

    Answer: d≈ cm
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(d)=−9πd2−2d2+42d=−(2+9π)d2+42d is a quadratic function. Notice that the quadratic coefficient, −(2+9π), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which d value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(d)=−9πd2−2d2+42dA′(d)=−18πd−4d+42
    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=−18πd−4d+420=−(4+18π)d+42⟵take out the d0=−(60,548...)d+42(60,548...)d=42d=42(60,548...)d≈0,69

    The value of d which maximizes the shaded area (rounded to 2 decimal places) is 0,69 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

    Submit your answer as:

  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 2 decimal places.

    Answer: height ≈ cm
    one-of
    type(numeric.noerror)
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 6d. We want the height which corresponds with the largest shaded area, so we need to use the value of d which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 6d. We also know from question (2) that the shaded area will be maximised when d=0,69. Therefore, we need to substitute this value into the expression for the height:

    heightmax=6d≈6(0,69)≈4,14

    When the shaded area is maximised, the height of the fake money (rounded to 2 decimal places) will be 4,14 cm.

    Submit your answer as:

Calculus applications: optimisation

Habubakar teaches Business Studies in the Western Cape. He wants to make fake money so that his students can use it to learn about business transactions. Habubakar decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of x, as labelled. The height of the rectangle will be 3x and the width is 8 cm.

  1. Find an expression for the shaded area of the fake money in terms of x.

    Answer: Shaded area A=
    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of x ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(3x)=3x2.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(8)×(3x)=(x)2=π(3x2)2=24x=x2=9π4x2
    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable x. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(x)shaded=Arectangle−2×Asquare−Acircle=24x−2(x2)−9π4x2=24x−2x2−9π4x2

    It is common to move the quadratic terms of x to the beginning of the function; so the shaded area is: A(x)=−9π4x2−2x2+24x.

    Submit your answer as:

  2. Determine the value of x for which the shaded area of the fake money will be maximised. Round your answer to 3 decimal places.

    Answer: x≈ cm
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?

    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(x)=−9π4x2−2x2+24x=−(2+9π4)x2+24x is a quadratic function. Notice that the quadratic coefficient, −(2+9π4), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which x value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(x)=−9π4x2−2x2+24xA′(x)=−9π2x−4x+24
    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=−9π2x−4x+240=−(4+9π2)x+24⟵take out the x0=−(18,137...)x+24(18,137...)x=24x=24(18,137...)x≈1,323

    The value of x which maximizes the shaded area (rounded to 3 decimal places) is 1,323 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.

    Submit your answer as:

  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 3 decimal places.

    Answer: height ≈ cm
    one-of
    type(numeric.noerror)
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 3x. We want the height which corresponds with the largest shaded area, so we need to use the value of x which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 3x. We also know from question (2) that the shaded area will be maximised when x=1,323. Therefore, we need to substitute this value into the expression for the height:

    heightmax=3x≈3(1,323)≈3,969

    When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be 3,969 cm.

    Submit your answer as:

Maximising the product of two unknown numbers

The sum of two positive numbers is 9. One of the numbers is multiplied by the square of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.

STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 9, and we are multiplying one of the numbers by the square of the other number.

Start by picking symbols for the two numbers. We will use a and b. Now use these variables to write equations:

The sum of the numbers is 9:

a+b=9…Eqn 1

One number multiplied by the square of the other:

P=ab2…Eqn 2

We use P to represent the product.

STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely a and b. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make a the subject and then substitute this expression into equation 2:

a+b=9∴a=9−b

Substituting:

P=ab2=(9−b)b2=9b2−b3=−b3+9b2

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.

STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=−b3+9b2∴dPdb=−3b2+18b
STEP: Find the value of b which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for b.

dPdb=−3b2+18b(0)=−3b2+18b

Divide through by a factor of −3:

0=b2−6b0=b(b−6)b=0orb=6

The solution b=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, b=6 corresponds to the maximum product.

STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of a which corresponds to this maximum, and then use both values to find the maximum product.

a=9−b=9−(6)=3

Finally:

P=ab2∴Pmax=(3)(6)2=(3)(36)=108

Therefore, the maximum value of the product is 108.

NOTE:There is a slightly different way to get the answer; once we have the value b=6, we can substitute it into the equation P=−b3+9b2 to get the maximum product. If we solve the question that way, we do not need to find the value of a at the maximum like we did above.
Submit your answer as:

Maximising the product of two unknown numbers

The sum of two positive numbers is 16. One of the numbers is multiplied by the cube of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.

STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 16, and we are multiplying one of the numbers by the cube of the other number.

Start by picking symbols for the two numbers. We will use a and b. Now use these variables to write equations:

The sum of the numbers is 16:

a+b=16…Eqn 1

One number multiplied by the cube of the other:

P=ab3…Eqn 2

We use P to represent the product.

STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely a and b. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make a the subject and then substitute this expression into equation 2:

a+b=16∴a=16−b

Substituting:

P=ab3=(16−b)b3=16b3−b4=−b4+16b3

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.

STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=−b4+16b3∴dPdb=−4b3+48b2
STEP: Find the value of b which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for b.

dPdb=−4b3+48b2(0)=−4b3+48b2

Divide through by a factor of −4:

0=b3−12b20=b2(b−12)b=0orb=12

The solution b=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, b=12 corresponds to the maximum product.

STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of a which corresponds to this maximum, and then use both values to find the maximum product.

a=16−b=16−(12)=4

Finally:

P=ab3∴Pmax=(4)(12)3=(4)(1728)=6912

Therefore, the maximum value of the product is 6912.

NOTE:There is a slightly different way to get the answer; once we have the value b=12, we can substitute it into the equation P=−b4+16b3 to get the maximum product. If we solve the question that way, we do not need to find the value of a at the maximum like we did above.
Submit your answer as:

Maximising the product of two unknown numbers

The sum of two positive numbers is 8. One of the numbers is multiplied by the cube of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.

STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 8, and we are multiplying one of the numbers by the cube of the other number.

Start by picking symbols for the two numbers. We will use m and n. Now use these variables to write equations:

The sum of the numbers is 8:

m+n=8…Eqn 1

One number multiplied by the cube of the other:

P=mn3…Eqn 2

We use P to represent the product.

STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely m and n. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make m the subject and then substitute this expression into equation 2:

m+n=8∴m=8−n

Substituting:

P=mn3=(8−n)n3=8n3−n4=−n4+8n3

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.

STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=−n4+8n3∴dPdn=−4n3+24n2
STEP: Find the value of n which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for n.

dPdn=−4n3+24n2(0)=−4n3+24n2

Divide through by a factor of −4:

0=n3−6n20=n2(n−6)n=0orn=6

The solution n=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, n=6 corresponds to the maximum product.

STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of m which corresponds to this maximum, and then use both values to find the maximum product.

m=8−n=8−(6)=2

Finally:

P=mn3∴Pmax=(2)(6)3=(2)(216)=432

Therefore, the maximum value of the product is 432.

NOTE:There is a slightly different way to get the answer; once we have the value n=6, we can substitute it into the equation P=−n4+8n3 to get the maximum product. If we solve the question that way, we do not need to find the value of m at the maximum like we did above.
Submit your answer as:

Calculus applications: optimisation

The Yummy in my Tummy food company makes dehaydrated beef powder (yum!) and the company wants to package the powder in small boxes, like the one shown below. The company will put 42 cubic centimetres of their food product into each box. The box has a base with a width of 4x cm, a depth of x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 4x cm, x cm, and h cm.

    Vbox=width×depth×height=(4x)(x)(h)=4hx2

    Therefore, the volume function for the box is 4hx2.

    Submit your answer as:

  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(4x)(h)+2(x)(h)+2(4x)(x)=10hx+8x2

    Therefore, the surface area equation for the box is 10hx+8x2.

    Submit your answer as:

  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 42 cm3. What can you do with that information?

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=4hx2 and surface area =10hx+8x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 42 cm3. Therefore, the equation for volume can be written as: 42=4hx2.

    Solve for h:

    42=4hx2424x2=h⇒h=212x2

    We have found that h=212x2.

    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=10hx+8x2=10(212x2)x+8x2=105x+8x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 8x2+105x.

    Submit your answer as:

  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.

    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=8x2+105x.

    It looks like the minimum surface area is close to 84 or 85. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=8x2+105xddx[S.A.(x)]=16x−105x2
    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=16x−105x2above by x2 to get this:multiply the equation→0=16x3−105105=16x310516=x3∴x=1,8722...

    The value of x which gives the minimum surface area is xmin≈1,87. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=8x2+105x=8(1,8722...)2+105(1,8722...)=8(3,5052...)+56,0832...=84,1248...2placesround to→≈84,12

    The minimum surface area for the box is 84,12 cm2. This answer also agrees with the graph above.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=212x2 and we also know that the width is simply w=4x. Use these two formulae to calculate the minimum values for each quantity:

    h=212x2w=4xhmin=212(1,8722...)2wmin=4(1,8722...)=2,9955...=7,4888...≈3,0≈7,49

    The height and width corresponding to the minimum surface area are h= 3,0 cm and w= 7,49 cm.

    Submit your answer as: andand

  5. If the cardboard used to make the boxes costs R0,006 per square centimetre, how much will it cost the company to produce 2 500 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 2 500 boxes is R .
    numeric
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 84,12 cm2 and the cardboard used to make the box costs R0,006 per cm2. How can you use these numbers to find the cost of one box?

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 2 500 boxes if each box requires 84,12 cm2 of cardboard. Given that that material for the boxes costs R0,006 per cm2, the total cost will be:

    Cost=(2 500 boxes)(R0,006cm2)(84,12 cm2box)=R1 261,80

    Therefore, the total cost for 2 500 boxes is R1 261,80.

    Submit your answer as:

Calculus applications: optimisation

The Nyama Nyama food company makes dried potato flakes and the company wants to package the flakes in small boxes, like the one shown below. The company will put 48 cubic centimetres of their food product into each box. The box has a base with a width of 4x cm, a depth of x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 4x cm, x cm, and h cm.

    Vbox=width×depth×height=(4x)(x)(h)=4hx2

    Therefore, the volume function for the box is 4hx2.

    Submit your answer as:

  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(4x)(h)+2(x)(h)+2(4x)(x)=10hx+8x2

    Therefore, the surface area equation for the box is 10hx+8x2.

    Submit your answer as:

  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 48 cm3. What can you do with that information?

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=4hx2 and surface area =10hx+8x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 48 cm3. Therefore, the equation for volume can be written as: 48=4hx2.

    Solve for h:

    48=4hx2484x2=h⇒h=12x2

    We have found that h=12x2.

    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=10hx+8x2=10(12x2)x+8x2=120x+8x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 8x2+120x.

    Submit your answer as:

  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.

    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=8x2+120x.

    It looks like the minimum surface area is close to 91 or 92. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=8x2+120xddx[S.A.(x)]=16x−120x2
    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=16x−120x2above by x2 to get this:multiply the equation→0=16x3−120120=16x312016=x3∴x=1,9574...

    The value of x which gives the minimum surface area is xmin≈1,96. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=8x2+120x=8(1,9574...)2+120(1,9574...)=8(3,8315...)+61,3047...=91,9571...2placesround to→≈91,96

    The minimum surface area for the box is 91,96 cm2. This answer also agrees with the graph above.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=12x2 and we also know that the width is simply w=4x. Use these two formulae to calculate the minimum values for each quantity:

    h=12x2w=4xhmin=12(1,9574...)2wmin=4(1,9574...)=3,1318...=7,8297...≈3,13≈7,83

    The height and width corresponding to the minimum surface area are h= 3,13 cm and w= 7,83 cm.

    Submit your answer as: andand

  5. If the cardboard used to make the boxes costs R0,003 per square centimetre, how much will it cost the company to produce 1 600 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 1 600 boxes is R .
    numeric
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 91,96 cm2 and the cardboard used to make the box costs R0,003 per cm2. How can you use these numbers to find the cost of one box?

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 1 600 boxes if each box requires 91,96 cm2 of cardboard. Given that that material for the boxes costs R0,003 per cm2, the total cost will be:

    Cost=(1 600 boxes)(R0,003cm2)(91,96 cm2box)=R441,41

    Therefore, the total cost for 1 600 boxes is R441,41.

    Submit your answer as:

Calculus applications: optimisation

The Yummy in my Tummy food company makes artificial butter flavouring for movie popcorn and the company wants to package the flavouring in small boxes, like the one shown below. The company will put 48 cubic centimetres of their food product into each box. The box has a base with a width of 4x cm, a depth of 2x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 4x cm, 2x cm, and h cm.

    Vbox=width×depth×height=(4x)(2x)(h)=8hx2

    Therefore, the volume function for the box is 8hx2.

    Submit your answer as:

  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?

    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(4x)(h)+2(2x)(h)+2(4x)(2x)=12hx+16x2

    Therefore, the surface area equation for the box is 12hx+16x2.

    Submit your answer as:

  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 48 cm3. What can you do with that information?

    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=8hx2 and surface area =12hx+16x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 48 cm3. Therefore, the equation for volume can be written as: 48=8hx2.

    Solve for h:

    48=8hx2488x2=h⇒h=6x2

    We have found that h=6x2.

    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=12hx+16x2=12(6x2)x+16x2=72x+16x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 16x2+72x.

    Submit your answer as:

  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.

    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=16x2+72x.

    It looks like the minimum surface area is close to 82 or 83. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=16x2+72xddx[S.A.(x)]=32x−72x2
    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=32x−72x2above by x2 to get this:multiply the equation→0=32x3−7272=32x37232=x3∴x=1,3103...

    The value of x which gives the minimum surface area is xmin≈1,31. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.

    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=16x2+72x=16(1,3103...)2+72(1,3103...)=16(1,7170...)+54,9462...=82,4194...2placesround to→≈82,42

    The minimum surface area for the box is 82,42 cm2. This answer also agrees with the graph above.

    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=6x2 and we also know that the width is simply w=4x. Use these two formulae to calculate the minimum values for each quantity:

    h=6x2w=4xhmin=6(1,3103...)2wmin=4(1,3103...)=3,4943...=5,2414...≈3,49≈5,24

    The height and width corresponding to the minimum surface area are h= 3,49 cm and w= 5,24 cm.

    Submit your answer as: andand

  5. If the cardboard used to make the boxes costs R0,003 per square centimetre, how much will it cost the company to produce 2 400 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 2 400 boxes is R .
    numeric
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 82,42 cm2 and the cardboard used to make the box costs R0,003 per cm2. How can you use these numbers to find the cost of one box?

    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 2 400 boxes if each box requires 82,42 cm2 of cardboard. Given that that material for the boxes costs R0,003 per cm2, the total cost will be:

    Cost=(2 400 boxes)(R0,003cm2)(82,42 cm2box)=R593,42

    Therefore, the total cost for 2 400 boxes is R593,42.

    Submit your answer as:

Rate of change: medicine in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain medicine in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=−t3+7t2+62t

where 0≤t≤12.

  1. Determine the number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken.

    Answer:

    The number of molecules of the medicine in the blood stream 5 minutes after the medicine was taken is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.

    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.

    M(t)=−t3+7t2+62tM(5)=−(5)3+7(5)2+62(5)=360

    The number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken is 360.

    Submit your answer as:

  2. Determine the rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken.

    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken is molecules per minute.

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.

    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=−t3+7t2+62tM′(t)=−3t2+14t+62
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M′(t)=−3t2+14t+62M′(2)=−3(2)2+14(2)+62=78

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken is 78 molecules per minute.

    Submit your answer as:

  3. How many minutes after taking the medicine will the rate at which the number of molecules of the medicine in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum after minute(s).

    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M′(t)=−3t2+14t+62M′′(t)=−6t+14
    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M′′(t)=−6t+140=−6t+1473=t

    The maximum rate of change of the number of molecules of the medicine in the bloodstream is after 73 minutes.

    Submit your answer as:

Rate of change: medicine in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain medicine in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=−t3+9t2+62t

where 0≤t≤13.

  1. Determine the number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken.

    Answer:

    The number of molecules of the medicine in the blood stream 5 minutes after the medicine was taken is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.

    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.

    M(t)=−t3+9t2+62tM(5)=−(5)3+9(5)2+62(5)=410

    The number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken is 410.

    Submit your answer as:

  2. Determine the rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken.

    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken is molecules per minute.

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.

    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=−t3+9t2+62tM′(t)=−3t2+18t+62
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M′(t)=−3t2+18t+62M′(4)=−3(4)2+18(4)+62=86

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken is 86 molecules per minute.

    Submit your answer as:

  3. How many minutes after taking the medicine will the rate at which the number of molecules of the medicine in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum after minute(s).

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M′(t)=−3t2+18t+62M′′(t)=−6t+18
    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M′′(t)=−6t+180=−6t+183=t

    The maximum rate of change of the number of molecules of the medicine in the bloodstream is after 3 minutes.

    Submit your answer as:

Rate of change: medicine in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain medicine in the bloodstream t hours after it had been taken is represented by the equation:

M(t)=−t3+4t2+64t

where 0≤t≤10.

  1. Determine the number of molecules of the medicine in the bloodstream 3 hours after the medicine was taken.

    Answer:

    The number of molecules of the medicine in the blood stream 3 hours after the medicine was taken is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of hours, 3, and evaluate.

    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of hours, 3, and evaluate.

    M(t)=−t3+4t2+64tM(3)=−(3)3+4(3)2+64(3)=201

    The number of molecules of the medicine in the bloodstream 3 hours after the medicine was taken is 201.

    Submit your answer as:

  2. Determine the rate at which the number of molecules of the medicine in the bloodstream is changing exactly 5 hours after the medicine was taken.

    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 5 hours after the medicine was taken is molecules per hour.

    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.

    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=−t3+4t2+64tM′(t)=−3t2+8t+64
    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M′(t)=−3t2+8t+64M′(5)=−3(5)2+8(5)+64=29

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 5 hours after the medicine was taken is 29 molecules per hour.

    Submit your answer as:

  3. How many hours after taking the medicine will the rate at which the number of molecules of the medicine in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum after hour(s).

    fraction
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M′(t)=−3t2+8t+64M′′(t)=−6t+8
    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M′′(t)=−6t+80=−6t+843=t

    The maximum rate of change of the number of molecules of the medicine in the bloodstream is after 43 hours.

    Submit your answer as: